Problem 36 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Center \((2,3),\) one focus \((0,3),\) one vertex (3,3)

Step-by-Step Solution

Answer

The hyperbola equation is \( (x-2)^2 - \frac{(y-3)^2}{3} = 1 \).

Step 1: Identify the Equation Type

A hyperbola with a horizontal transverse axis has the general equation: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center.

Step 2: Determine the Center

Given that the center of the hyperbola is \((2,3)\), we know \(h = 2\) and \(k = 3\). So, the equation begins as: \( \frac{(x-2)^2}{a^2} - \frac{(y-3)^2}{b^2} = 1 \).

Step 3: Calculate 'a', the Distance from Center to Vertex

The vertex is given as \((3,3)\), which implies that the vertex lies one unit to the right of the center. Thus, \(a = 1\).

Step 4: Find 'c', the Distance from Center to Focus

Since one focus is \((0,3)\), the distance from the center \((2,3)\) to this focus is \(c = 2\).

Step 5: Calculate 'b' Using the Relationship

For hyperbolas, the relationship is \( c^2 = a^2 + b^2 \). Substituting the known values, we have \(2^2 = 1^2 + b^2\). Thus, \(4 = 1 + b^2\), which gives \(b^2 = 3\).

Step 6: Write the Final Equation

Substitute \(a^2 = 1\) and \(b^2 = 3\) into the hyperbola equation. The final equation is: \( \frac{(x-2)^2}{1} - \frac{(y-3)^2}{3} = 1 \), or simplified as \( (x-2)^2 - \frac{(y-3)^2}{3} = 1 \).

Key Concepts

Equation of Hyperbola

The equation of a hyperbola plays a crucial role in defining its shape and orientation. The standard form of a hyperbola's equation with a horizontal transverse axis is:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]In this equation:

• \((h, k)\) represents the center of the hyperbola
• \(a\) is the distance from the center to each vertex along the transverse axis
• \(b\) relates to the distance along the conjugate axis, which does not determine the location of real parts like vertices and foci

The given exercise asks to find the equation of a hyperbola when certain geometric points are specified. In our example, with the center provided at \((2,3)\), one vertex at \((3,3)\), and one focus at \((0,3)\), we can follow the steps to determine the equation as:\[ (x-2)^2 - \frac{(y-3)^2}{3} = 1 \]This equation tells us about the hyperbola's orientation and dimensions.

Center of Hyperbola

The center of a hyperbola is the midpoint between its vertices and is also equidistant from its foci. In a coordinate plane, the center helps in determining the pivotal point around which the hyperbola is oriented.For the given problem, the center was provided at \((2,3)\). Knowing the center plays a key role in writing the hyperbola's equation. The coordinates \((h, k)\) are plugged into the hyperbola equation so that:\[ \frac{(x-2)^2}{a^2} - \frac{(y-3)^2}{b^2} = 1 \]Here,

• \(h = 2\) and \(k = 3\) help define the equation's location on the plane
• Everything else revolves around this reference point

Understanding where the center is located guides both the construction and visual representation of the hyperbola.

Vertex and Focus of Hyperbola

Vertices and foci are pivotal elements in the structure of a hyperbola.- **Vertex:** A vertex of a hyperbola is a point where it makes its sharpest turn. The vertices lie along the transverse axis, defining the real AND visible 'arms' of the hyperbola. In the problem, a vertex is provided at \((3,3)\). This means it is 1 unit from the center \((2,3)\). Thus, we identify \(a = 1\).- **Focus:** The focus of a hyperbola is a point located inside each 'arm'. It helps determine the curve's extent of deviation. For this exercise, a focus is given at \((0,3)\). The distance from the center \((2,3)\) to this focus is \(c = 2\).Both these elements are used to further solve or establish the hyperbola's shape using the formula:\[ c^2 = a^2 + b^2 \]Substituting the known values, we found \(b^2 = 3\). Together, these elements contribute to the hyperbola's overall equation and give insight into its graphical representation.